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Supposing we want to formulate a very primitive theory of integration, the only requirement being that all continuous functions $[a, b]\longrightarrow\mathbb{R}$ be integrable. What is the simplest possible theory that includes a rigorous notion of integral and a proof that it exists for all continuous functions?

Clearly the Riemann-integral gives way to the simpler regulated integral defined via uniform limits of step functions, but it still takes a little non-trivial work to show that this includes the continuous functions. A candidate to beat this is to just to take a Riemann sum with equidistant partition

$$\int_a^b f(x)dx:=\lim\limits_{n\longrightarrow\infty}\frac{b-a}{n}\sum\limits_{k=1}^n f\left(a+\frac{k}{n}(b-a)\right)$$

which boils down to showing that this limit exists if $f$ is continuous, but this still requires some educated feeling for how to proceed since we can't avoid using an instance of the fact that cotinuous functions on $[a, b]$ are uniformly continuous. I am wondering if we can beat this... Is there some natural elementary way to rigorously introduce a primitive integral, for which the proof of continuous being integrable sort of writes itself?

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  • $\begingroup$ This is a very good question, in my opinion. Just as an aside: there are other notions of integration besides Riemann - and even Lebesgue. They aren't as well known but at least one of them bridges the gap between Riemann and Lebesgue quite nicely from what I recall. Particularly, it allows for a nice theory of improper integrals while still handling Lebesgue-type theory. $\endgroup$ Aug 1, 2015 at 19:51
  • $\begingroup$ You must be speaking of the Henstock-Kurzweil gauge integral - I appreciate its merits, but it requires fine tagged partitions or Cousin-covers to introduce, so as far as simplicity is concerned it is on the other extreme end :-) $\endgroup$ Aug 1, 2015 at 19:54
  • $\begingroup$ Henstock is the one. I couldn't remember the name. There are good reasons for why the others aren't as well known and, as you say, they are quite a bit more difficult to build up than either the Riemann or Lebesgue integrals. I was just throwing it out there since not many people know of it and it's pretty neat although very dense. $\endgroup$ Aug 1, 2015 at 19:56
  • $\begingroup$ I think the reason is that the construction is specific to the real line and it doesn't generalize easily to abstract context like Lebesgue does. $\endgroup$ Aug 1, 2015 at 19:57
  • $\begingroup$ If the function $f$ is bounded and continuous a.e. on $[a,b]$, then $f$ is Riemann integrable, and any such sequence of sums over equally-spaced partitions will have to converge to the Riemann integral. $\endgroup$ Aug 1, 2015 at 20:12

2 Answers 2

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For those interested, I may now have found an answer, which is quite satisfactory in the above sense and perhaps of some value in the teaching of elementary analysis:

  1. Start with Weierstrass' Approximation Theorem: A continuous function $f$: $[0, 1]\longrightarrow\mathbb{R}$ is the uniform limit of polynomials $(p_n)$. Admittedly, this is not trivial to prove, but it can be convincingly used without proof or proved nicely by writing down the concrete sequence of Bernstein polynomials, and the proof also isolates away uniform continuity arguments so they don't need to be invoked in further construction.

  2. The obvious idea to set $\int f :=\lim_n\int p_n$ is no good as it requires us to evaluate integrals of powers $x^m$ via Riemann sums (s. above formula) - this isn't easy at all, but there is a nice way round it via the exponential function: Note that if $p_n$ converges uniformly to $f$, then $p_n\circ\exp$ converges uniformly to $f\circ\exp$ - this draws upon the fact that a continuous bijection on a compact interval has a continuous inverse. So continuous functions on $[0, 1]$ are uniform limits of polynomials composed with $\exp$, or consequently, linear combinations of exp-functions.

  3. The above boils it all down to the computation of integrals of $e^{cx}$ (some $c>0$) via Riemann sums, but this is easy since $\sum_{k=0}^{n-1} e^{ck}=\frac{1-e^{cn}}{1-e^c}$ by telescoping and all one needs is the knowledge that $\lim_{x\longrightarrow 0}\frac{e^x-1}{x}=1$, which is provable by simple algebraic manipulation.

  4. Extension to $[a, b]$ from $[0, 1]$ is easily done by linear substitution.

  5. This notion of integral (which so far works for continuous functions only but is good enough for a primitive version of the Fundamental Theorem of Calculus) can then be further extended in the obvious way to piecewise continuous functions and then subsequently to uniform limits of such. The latter in fact turn out to be the precisely the regulated functions (but that's where some Heine-Borel type compactness argument is required at the latest).

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  • $\begingroup$ I like this way, but how can you explain simply that the integral of a continuous function is just the algebraic area ? $\endgroup$
    – Hamza
    Aug 15, 2015 at 21:19
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    $\begingroup$ Well, it's clearly the area for the base functions $e^{cx}$ because that's how we computed the integral (sum of rectangles of equal length), then the integral of a continuous functions is the uniform limit of linear combinations of the base functions, so it should be believable enough to a beginner. If you want to do it rigorously, I'm not sure about an easy way.. maybe by using the intermediate value theorem to show that the area $A=\int f$ is attained by some rectangle A=(b-a)f(x_0) (which essentially leads to the mean value theorem for integration). $\endgroup$ Aug 15, 2015 at 21:30
  • $\begingroup$ Very clever and satisfying. Taking your axioms in the right way, the integral of $e^x$ is trivial. Great idea. $+1$ $\endgroup$ Aug 15, 2015 at 21:39
  • $\begingroup$ I have the (uneducated) feeling that a pedagogically optimized first course on analysis should be using a primitive integral along these lines, just to illustrate the statement of FTC for a simple class of functions, instead of mixing it up with Riemann integration, which the average beginner does not understand. $\endgroup$ Aug 15, 2015 at 21:46
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    $\begingroup$ @mathgemini This is quite an interesting approach. But I don't believe that this way forward is simpler than RI theory. Step functions and uniform continuity seem, in my humble opinion, much simpler to teach/learn than the WAT, and invoking uniform convergence ideas. So, a BIG +1 for the question and another for the answer. But, ... just don't agree on the notion of simplicity. Of course, you might have not meant simpler. Good work. $\endgroup$
    – Mark Viola
    Aug 15, 2015 at 22:27
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This question just popped up again from an edit and I was rather surprised that, considering the amount of interest, no one has suggested a rather obvious integration theory that also answers the question.

We want a "teaching integral" that is adequate to get the students started on a study of integration theory but not overwhelm them with the correct modern theories. Some do suggest the Henstock-Kurzweil integral. This makes sense only if the students have already studied the rather pathetic Riemann integral. We say, then, "Here is a better theory, no harder than what you just learned, and much more powerful if you care to examine it further." But the Henstock-Kurzweil integral is a solution to a different problem. It is not "integration with training wheels." It is a full theory of integration that includes the Riemann integral, the Lebesgue integral, the improper Riemann integral, and the Denjoy-Perron integral.

Some of us feel that a return to the Newton integral is a better introduction to integration theory as a first step. After all, this is the way that every 18th century mathematician viewed integration theory.

Definition A function $f:[a,b]\to\mathbb R$ is Newton integrable on $[a,b]$ if there is an antiderivative $F$ (i.e., $F'(x)=f(x)$ for all $a\leq x \leq b$) and then one defines $$\int_a^b f(x)\,dx=F(b)-F(a).$$

The justification for the integral is simply the mean-value of the calculus. The basic properties of the integral are all deduced from properties of derivatives. Riemann sums come in by way of the mean-value theorem too.

I think if you did a poll of calculus students who have been force-fed the Riemann integral, nearly all of them would say that this is indeed all that they consider integration to be. It is how they all calculate integrals --although they remember vaguely some unpleasant times spent using Riemann sums, but mercifully no-one makes them do that stuff any more.

Does this integral handle all continuous functions? You can do exactly as Cauchy did and show how to construct the primitive of continuous functions using Riemann sums. I prefer myself to prove this lemma:

Lemma Let $f:[a,b]\to\mathbb R$ be a bounded function. Then there is a Lipschitz function $F:[a,b]\to\mathbb R$ so that $F'(x)=f(x)$ at every point $x$ at which $f$ is continuous.

It is not that hard and can be presented at an elementary level if you don't demand the students follow in detail every step. You don't need uniform continuity evidently, but you do need to construct a sequence of functions that converges to the function you want.

For more information about how one might possibly take this point of view for an introductory course in integration theory you can consult this experimental textbook.

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  • $\begingroup$ Thanks. I appreciate the lemma, but not sure about the Newton integral. While I can live with an integral notion that provides no intuition as to why the value is the area under the curve, I'm not sure if I can live with an integral notion that trivializes the Fundamental Theorem of Calculus..... or am I getting something terribly wrong? $\endgroup$ Jan 20, 2016 at 21:06
  • $\begingroup$ You are being unfair ... not to me, to Newton. Once a student understands differentiation (fluxions) we can explain Newton's great insight as to why the ancient problem of quadrature (i.e., finding area under a curve) can be solved by the reverse process. And it is easy enough to explain that now. Integration as the reverse process of differentiation dominated intellectual thought for a long time. The FTC as it is taught in calculus is entirely ahistorical. What's wrong with starting with Newton's integral as a prelude to modern integration theory? $\endgroup$ Jan 20, 2016 at 21:34
  • $\begingroup$ ... (cont.) Even the Lebesgue integral can be realized as a Newton-type integral (as can the Denjoy integral). You relax $F'=f$ to be a.e. and impose a stronger condition on $F$. The nexus between integration as reverse differentiation and integration as area is a dominant theme in our subject. Even in general measure theory in abstract setttings this theme can be developed. The FTC doesn't trivialize, it just shifts in meaning. This is a legitimate point of view, but not one that will appeal to all instructors of the subject for a first course. But the Riemann integral is a worse choice. $\endgroup$ Jan 20, 2016 at 21:37
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    $\begingroup$ For the Ark? Maybe I would vote for the "regulated integral." Step functions + uniform limits seems compelling enough. But you get only one female companion I presume? If you select her wisely enough she would be happy with a copy of Federer's Geometric Measure Theory to bring our theories to a new post-diluvial world. $\endgroup$ Jan 21, 2016 at 17:13
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    $\begingroup$ To be honest I'd rather select a female companion who would be happy with a copy of a survey of the most beautiful and elegant results and constructions concerning functions of one real variable, much in the style of Aigner&Ziegler's "Proofs from THE BOOK". $\endgroup$ Jan 21, 2016 at 21:48

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